Prove: There exists exactly one prime triple. (No others exists),?

Question

Definition: a prime triple is a set of three consecutive odd numbers, each of which is a prime.

also prove these:
Prove: If c is an odd integer then n square + n - c = 0 has no integer solutions.

For real number a and b, prove: If a

Answer 1

The first one looks hard at first, but it's pretty easy.

Consecutive odd numbers can be expressed as n, n+2, and n+4, as long as n is odd.

All values of n can be split into n/3, (n+1)/3, and (n+2)/3, in which each expression is also an integer. If n/3 is an integer, then obviously n is composite. If (n+1)/3 is an integer, then n+4 must also be divisible by 3, and therefore composite. If (n+2)/3 is an integer, then obviously n+2 is composite.

There are only 3 flaws in this proof. One is when n=1, in which you get 1, 3, 5. 1 is not prime by definition. Another is when n=3, and there you have your only triple; 3, 5, 7. The last one is when n=-1, but negative numbers are always composite as well.

Here's the second one.

If n is even, n squared is even, and n squared plus n is even. Therefore, c must be even for there to be an integer solution.

If n is odd, n sqaured is odd, and n squared plus n is again even. Therefore c must be even again for there to be integer solution.

These work as long as n is rational. I'm not sure about irrational, etc.

I can also prove the third one. All you are saying is that the average of two numbers is always between them. I'll rewrite the problem as...

2a/2<(a+b)/2<2b/2

Obviously, 2a/2=a and 2b=b.

Since a Since b>a, b+b>b+a or 2b>a+b, and 2b/2>(a+b)/2.
Rewrite as 2a/2<(a+b)/2<2b/2 or a<(a+b)/2

Answer 2

We see (3,5,7) is a prime triple. So, there exists one. If (m, m+2, m+4) is a different prime triple, then m>3, so that all the 3 terms are greater than 3.

Every positive odd number n is of the form n = 3*p + k, where p is a positive odd number and k is in {0, 1, 2} Therefore, 3 consecutive odd numbers are of the form 3*p + k, 3*p + k+1, 3*p + k+2. Out of the numbers, k, k+1 , k+2, one and only one is multiple of 3, which implies out of 3 consecutive positive odd numbers, one and only one is multiple of 3.

This implies that, If (m, m+2, m+4) is a prime triple distinct from (3,5, 7), then one of its terms is greater than 3 and is divisible by 3. Therefore this term has a divisor different from 1 and from itself, contrarily to the assumption (m, m+2, m+4) is a prime triple. This proves (3, 5 , 7) is the only prime triple


The second assertion follows from a therorem which says polynomials of degree 2 with integer odd coefficients have no rational roots.

If a < b, the (a + b)/2 > (a + a)/ 2= a and (a +b )/2 < (b + b)/2 = b.

That's it.

Answer 3

if there are 3 conscutive odd numbers
they are p, p+2 p+4
p can be of the form 3n, 3n+1, 3n+2
if p is multiple of 3 that is 3n it can be prime only if n = 1 then we get 3,5,7 which are prime.
if n !=1 then p is not a prime.

if p is of the form 3n+1 then p+2 = 3(n+1) is not prime as 2 factors 3 and n+1

if p is of the form 3n +2 then p+4 = 3(n+2) is not a prime.

so there is only one prime tripple.

2)
n^2+n = c
=> n(n+1) = c
either n or n+1 is even that is both cannot be odd . so c = odd does not have integral solution
3)
a < b
add a on both sides
a + a < b +a ...or 2a < b+ a or a < (b+a)/2 ...1
add b on both sides
a+b < 2b or (b+a)/2 < b ....2

from 1 and 2 we get a <(b+a)/2 < b

Answer 4

ok, so consider a triplet of consequtive odd numbers. If the first one is prime then it is either 1 or 2 modulo 3. Add 2 to that and you get either 0 or 1 modulo 3. If you got 0 then you're done, if you got 1 then add 2 to that and your back to 0 mod 3. The point here is that any time you have 3 consequtive odd numbers, one of them has to be 0 mod 3.